solution to system of equations

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In a system of equations AX = B.
by Cramer's rule if det(A)=0, then the system will have no solution but on the other hand, in non-homogeneous system of equations suppose the rank of square matrix nxn is (n-1) and rank of the augmented matrix[A|B] is also having the same rank but then we say that system is consistent, having rank < no. of variables implies infinitely many solutions.

how is this correct?
I mean if the rank of A is n-1 < n and rank(A) = rank(A|B), then it also implies that determinant of square matrix nxn is zero but then we say it is consistent.

pls correct me if my theory is wrong too.

linear-algebra systems-of-equations

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edited Aug 15 at 12:43

Harry Peter

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asked Aug 10 at 5:12

ashwani yadav

234

add a comment | 

up vote
2
down vote

favorite

In a system of equations AX = B.
by Cramer's rule if det(A)=0, then the system will have no solution but on the other hand, in non-homogeneous system of equations suppose the rank of square matrix nxn is (n-1) and rank of the augmented matrix[A|B] is also having the same rank but then we say that system is consistent, having rank < no. of variables implies infinitely many solutions.

how is this correct?
I mean if the rank of A is n-1 < n and rank(A) = rank(A|B), then it also implies that determinant of square matrix nxn is zero but then we say it is consistent.

pls correct me if my theory is wrong too.

linear-algebra systems-of-equations

share|cite|improve this question

edited Aug 15 at 12:43

Harry Peter

5,49311438

asked Aug 10 at 5:12

ashwani yadav

234

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

In a system of equations AX = B.
by Cramer's rule if det(A)=0, then the system will have no solution but on the other hand, in non-homogeneous system of equations suppose the rank of square matrix nxn is (n-1) and rank of the augmented matrix[A|B] is also having the same rank but then we say that system is consistent, having rank < no. of variables implies infinitely many solutions.

how is this correct?
I mean if the rank of A is n-1 < n and rank(A) = rank(A|B), then it also implies that determinant of square matrix nxn is zero but then we say it is consistent.

pls correct me if my theory is wrong too.

linear-algebra systems-of-equations

share|cite|improve this question

edited Aug 15 at 12:43

Harry Peter

5,49311438

asked Aug 10 at 5:12

ashwani yadav

234

In a system of equations AX = B.
by Cramer's rule if det(A)=0, then the system will have no solution but on the other hand, in non-homogeneous system of equations suppose the rank of square matrix nxn is (n-1) and rank of the augmented matrix[A|B] is also having the same rank but then we say that system is consistent, having rank < no. of variables implies infinitely many solutions.

how is this correct?
I mean if the rank of A is n-1 < n and rank(A) = rank(A|B), then it also implies that determinant of square matrix nxn is zero but then we say it is consistent.

pls correct me if my theory is wrong too.

linear-algebra systems-of-equations

share|cite|improve this question

edited Aug 15 at 12:43

Harry Peter

5,49311438

asked Aug 10 at 5:12

ashwani yadav

234

share|cite|improve this question

share|cite|improve this question

edited Aug 15 at 12:43

Harry Peter

5,49311438

edited Aug 15 at 12:43

Harry Peter

5,49311438

edited Aug 15 at 12:43

Harry Peter

5,49311438

5,49311438

asked Aug 10 at 5:12

ashwani yadav

234

asked Aug 10 at 5:12

ashwani yadav

234

asked Aug 10 at 5:12

ashwani yadav

234

234

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
3
down vote



accepted

Cramer rule doesn't says that when determinant is zero, it has no solution. When the determinant is zero, the system can either have no solution (inconsistent) or infinitely many solutions (consistent).

Cramer rule gives us a way to compute the solution when the solution is unique.

share|cite|improve this answer

answered Aug 10 at 5:13

Siong Thye Goh

80.5k1453101

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks , doing a question got mistaken somewhere
  – ashwani yadav
  Aug 10 at 5:19
  

  
  
  
  

add a comment | 

up vote
3
down vote

For the system of two equations:

$ax+by=c$ and $Ax+By=C$

compute $D=beginvmatrixa&b\A&Bendvmatrix$, $D_x=beginvmatrixc&b\C&Bendvmatrix$ and $D_y=beginvmatrixa&c\A&Cendvmatrix$. Now you have two cases:

1. If $Dne 0$ then the unique solution is $x=fracD_xD$ and $y=fracD_yD$.


2. If $D=0$ then Cramer's rule fails to give the solution (if it exists).

share|cite|improve this answer

edited Aug 10 at 6:14

answered Aug 10 at 6:03

Mathlover

3,5681021

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since when could you divide matrices?
  – Chase Ryan Taylor
  Aug 10 at 6:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh sorry! I used wrong brackets for determinats.
  – Mathlover
  Aug 10 at 6:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $ddotsmile$
  – Chase Ryan Taylor
  Aug 10 at 6:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Edited it @Chase Ryan Taylor. Thanks a lot!
  – Mathlover
  Aug 10 at 6:15
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

Cramer rule doesn't says that when determinant is zero, it has no solution. When the determinant is zero, the system can either have no solution (inconsistent) or infinitely many solutions (consistent).

Cramer rule gives us a way to compute the solution when the solution is unique.

share|cite|improve this answer

answered Aug 10 at 5:13

Siong Thye Goh

80.5k1453101

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks , doing a question got mistaken somewhere
  – ashwani yadav
  Aug 10 at 5:19
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

Cramer rule doesn't says that when determinant is zero, it has no solution. When the determinant is zero, the system can either have no solution (inconsistent) or infinitely many solutions (consistent).

Cramer rule gives us a way to compute the solution when the solution is unique.

share|cite|improve this answer

answered Aug 10 at 5:13

Siong Thye Goh

80.5k1453101

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks , doing a question got mistaken somewhere
  – ashwani yadav
  Aug 10 at 5:19
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

Cramer rule doesn't says that when determinant is zero, it has no solution. When the determinant is zero, the system can either have no solution (inconsistent) or infinitely many solutions (consistent).

Cramer rule gives us a way to compute the solution when the solution is unique.

share|cite|improve this answer

answered Aug 10 at 5:13

Siong Thye Goh

80.5k1453101

Cramer rule doesn't says that when determinant is zero, it has no solution. When the determinant is zero, the system can either have no solution (inconsistent) or infinitely many solutions (consistent).

Cramer rule gives us a way to compute the solution when the solution is unique.

share|cite|improve this answer

answered Aug 10 at 5:13

Siong Thye Goh

80.5k1453101

share|cite|improve this answer

share|cite|improve this answer

answered Aug 10 at 5:13

Siong Thye Goh

80.5k1453101

answered Aug 10 at 5:13

Siong Thye Goh

80.5k1453101

answered Aug 10 at 5:13

Siong Thye Goh

80.5k1453101

80.5k1453101

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks , doing a question got mistaken somewhere
  – ashwani yadav
  Aug 10 at 5:19
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks , doing a question got mistaken somewhere
  – ashwani yadav
  Aug 10 at 5:19
  

  
  
  
  

thanks , doing a question got mistaken somewhere
– ashwani yadav
Aug 10 at 5:19

thanks , doing a question got mistaken somewhere
– ashwani yadav
Aug 10 at 5:19

add a comment | 

up vote
3
down vote

For the system of two equations:

$ax+by=c$ and $Ax+By=C$

compute $D=beginvmatrixa&b\A&Bendvmatrix$, $D_x=beginvmatrixc&b\C&Bendvmatrix$ and $D_y=beginvmatrixa&c\A&Cendvmatrix$. Now you have two cases:

1. If $Dne 0$ then the unique solution is $x=fracD_xD$ and $y=fracD_yD$.


2. If $D=0$ then Cramer's rule fails to give the solution (if it exists).

share|cite|improve this answer

edited Aug 10 at 6:14

answered Aug 10 at 6:03

Mathlover

3,5681021

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since when could you divide matrices?
  – Chase Ryan Taylor
  Aug 10 at 6:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh sorry! I used wrong brackets for determinats.
  – Mathlover
  Aug 10 at 6:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $ddotsmile$
  – Chase Ryan Taylor
  Aug 10 at 6:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Edited it @Chase Ryan Taylor. Thanks a lot!
  – Mathlover
  Aug 10 at 6:15
  

  
  
  
  

add a comment | 

up vote
3
down vote

For the system of two equations:

$ax+by=c$ and $Ax+By=C$

compute $D=beginvmatrixa&b\A&Bendvmatrix$, $D_x=beginvmatrixc&b\C&Bendvmatrix$ and $D_y=beginvmatrixa&c\A&Cendvmatrix$. Now you have two cases:

1. If $Dne 0$ then the unique solution is $x=fracD_xD$ and $y=fracD_yD$.


2. If $D=0$ then Cramer's rule fails to give the solution (if it exists).

share|cite|improve this answer

edited Aug 10 at 6:14

answered Aug 10 at 6:03

Mathlover

3,5681021

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since when could you divide matrices?
  – Chase Ryan Taylor
  Aug 10 at 6:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh sorry! I used wrong brackets for determinats.
  – Mathlover
  Aug 10 at 6:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $ddotsmile$
  – Chase Ryan Taylor
  Aug 10 at 6:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Edited it @Chase Ryan Taylor. Thanks a lot!
  – Mathlover
  Aug 10 at 6:15
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

For the system of two equations:

$ax+by=c$ and $Ax+By=C$

compute $D=beginvmatrixa&b\A&Bendvmatrix$, $D_x=beginvmatrixc&b\C&Bendvmatrix$ and $D_y=beginvmatrixa&c\A&Cendvmatrix$. Now you have two cases:

1. If $Dne 0$ then the unique solution is $x=fracD_xD$ and $y=fracD_yD$.


2. If $D=0$ then Cramer's rule fails to give the solution (if it exists).

share|cite|improve this answer

edited Aug 10 at 6:14

answered Aug 10 at 6:03

Mathlover

3,5681021

For the system of two equations:

$ax+by=c$ and $Ax+By=C$

compute $D=beginvmatrixa&b\A&Bendvmatrix$, $D_x=beginvmatrixc&b\C&Bendvmatrix$ and $D_y=beginvmatrixa&c\A&Cendvmatrix$. Now you have two cases:

1. If $Dne 0$ then the unique solution is $x=fracD_xD$ and $y=fracD_yD$.


2. If $D=0$ then Cramer's rule fails to give the solution (if it exists).

share|cite|improve this answer

edited Aug 10 at 6:14

answered Aug 10 at 6:03

Mathlover

3,5681021

share|cite|improve this answer

share|cite|improve this answer

edited Aug 10 at 6:14

edited Aug 10 at 6:14

edited Aug 10 at 6:14

answered Aug 10 at 6:03

Mathlover

3,5681021

answered Aug 10 at 6:03

Mathlover

3,5681021

answered Aug 10 at 6:03

Mathlover

3,5681021

3,5681021

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since when could you divide matrices?
  – Chase Ryan Taylor
  Aug 10 at 6:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh sorry! I used wrong brackets for determinats.
  – Mathlover
  Aug 10 at 6:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $ddotsmile$
  – Chase Ryan Taylor
  Aug 10 at 6:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Edited it @Chase Ryan Taylor. Thanks a lot!
  – Mathlover
  Aug 10 at 6:15
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since when could you divide matrices?
  – Chase Ryan Taylor
  Aug 10 at 6:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh sorry! I used wrong brackets for determinats.
  – Mathlover
  Aug 10 at 6:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $ddotsmile$
  – Chase Ryan Taylor
  Aug 10 at 6:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Edited it @Chase Ryan Taylor. Thanks a lot!
  – Mathlover
  Aug 10 at 6:15
  

  
  
  
  

Since when could you divide matrices?
– Chase Ryan Taylor
Aug 10 at 6:12

Since when could you divide matrices?
– Chase Ryan Taylor
Aug 10 at 6:12

Oh sorry! I used wrong brackets for determinats.
– Mathlover
Aug 10 at 6:13

Oh sorry! I used wrong brackets for determinats.
– Mathlover
Aug 10 at 6:13

$ddotsmile$
– Chase Ryan Taylor
Aug 10 at 6:14

$ddotsmile$
– Chase Ryan Taylor
Aug 10 at 6:14

Edited it @Chase Ryan Taylor. Thanks a lot!
– Mathlover
Aug 10 at 6:15

Edited it @Chase Ryan Taylor. Thanks a lot!
– Mathlover
Aug 10 at 6:15

add a comment | 

 

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